Question

The spontaneous reaction that takes place in this cell is:

• A. $Zn+Ni\to Z{n}^{2+}+N{i}^{2+}$
• B. $Zn+N{i}^{2+}\to Z{n}^{2+}+Ni$
• C. $Ni+Z{n}^{2+}\to N{i}^{2+}+Zn$
• D. $N{i}^{2+}+2K\to 2K^{+}+Ni$

Answer 1

The correct option is B $Zn+N{i}^{2+}\to Z{n}^{2+}+Ni$

$Zn$ is at lower potential, it will be oxidised and the cell reaction will be:

$Z{n}^{2+}+2e^{-}\rightleftharpoons Zn$ $E^0=-0.76V$

$N{i}^{2+}+2e^{-}\rightleftharpoons Ni$ $E^0=-0.23V$

$Zn\left(s\right)+N{i}^{2+}\left(aq\right)\to Z{n}^{2+}(aq.)+Ni\left(s\right)$.

$E_{cell}^0=0.76-0.23=0.53V$

Hence, option B is correct.